Problem Statement
Consider the expansion of $\left(2x^2 - \frac{1}{x}\right)^9$.
**(a)** [3 marks] Find the value of $r$ corresponding to the term independent of $x$ (the constant term).
**(b)** [2 marks] Hence, evaluate the constant term in exact integer form.
Verified Solution & Marking Scheme
Part (a) [M1]
General term: $T_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left(-\frac{1}{x}\right)^r = \binom{9}{r} 2^{9-r} (-1)^r x^{2(9-r) - r}$
Part (a) [M1]
Power of $x$ is $18 - 2r - r = 18 - 3r$. Set $18 - 3r = 0$
Part (a) [A1]
$3r = 18 \implies r = 6$
Part (b) [M1]
Substitute $r = 6$: $T_7 = \binom{9}{6} 2^{9-6} (-1)^6 = \binom{9}{3} 2^3 (1)$
Part (b) [A1]
$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \implies T_7 = 84 \times 8 = 672$